
The angular part is a standard integral [1]},
and \(I_{PT_2}\)  becomes
\(\begin{split}I_{PT_2}= &-i \pi ^2 m \int _0^\Lambda \!\! dr\frac{r (r^2+m^2-2g^2f_1)}{g^2\sqrt{f_1^2+16f_2^2}}\ln \Big [1-\frac{B(r)}{A(r)}\Big ]\end{split}\) 
